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Chapter 9: Chemical Equilibrium

Chapter 9: Chemical Equilibrium. Homework: Exercises (a only): 9.5, 6, 7, 9, 11, 12 Problems: 9.1, 8. Reaction Gibbs Energy. Consider the simple reaction A ˛ B Examples d-alanine to l- alanine Suppose an infintesimal amount of n A converts to n B , then dn A =-dn B

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Chapter 9: Chemical Equilibrium

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  1. Chapter 9: Chemical Equilibrium Homework: Exercises (a only): 9.5, 6, 7, 9, 11, 12 Problems: 9.1, 8

  2. Reaction Gibbs Energy • Consider the simple reaction A ˛ B • Examples d-alanine to l- alanine • Suppose an infintesimal amount of nA converts to nB, then dnA =-dnB • If we define x (ksi)as the extent of reaction then d x = -dnA =dnB • Dimensions of x are moles • For finite reaction the change is Dx and nA goes to nA - Dx and nB goes to nB +Dx • Reaction Gibbs Energy is the slope of the Gibbs energy vs. x or DrG = (dG/d x) p,T • Remember, dG = µAdnA+ µBdnBso dG = -µAd x+ µBd x or DrG = (dG/d x ) p,T= µB -µA • Reaction Gibbs energy is the difference of the product chemical potential and the reactant chemical potential

  3. Gibbs Energy and Equilibrium • Initially µB <µA so DrG <0 and the reaction is spontaneous • Called an exergonic reaction (work producing) • If µB >µA so DrG >0 and the reverse reaction is spontaneous • Called an endergonic reaction (work consuming) • When µB =µA so DrG >0 or system is at equilibrium • Thus a minimum in the reaction Gibbs energy plot represents equilibrium • At a minimum, the slope = 0

  4. Ideal Gas Equilibria • Consider homogenous gas phase reactions • Reactants and products are gases • Gases to 1st approximation behave as ideal gases • Since DrG = (dG/d x ) p,T = µB -µA and µ = µ° + RT ln(p), DrG = µB° + RT ln(p B) - (µA°+ RT ln(p A)) = µB°- µA° + RT ln(pB /pA) DrG = DrG° + RT ln(Q) where • DrG° = µB° - µA°, the standard reaction Gibbs energy • Like std reaction enthalpy it is the difference in standard free energies of formation of the products - that of reactants • Q = pB /pA, the reaction quotient which ranges from 0 (pure A) to infinity (pure B) • At equilibrium, DrG = 0, so DrG° = - RT ln(K) or K = exp(-DrG°/RT) • K is the equilibrium value of Q or K = (pB /pA )equilibrium

  5. Comments on the Above • The minimum in Gibbs energy arises from the mixing of reactant and products • If there were no mixing, G would change linearly in proportion to the amount of B formed • Slope of G vs. x would be DrG° • In chapter 7, we learned DmixG° = nRT(x A ln(x A) + x B ln(x B) whi9ch is a U shaped function (minimum at 50% B) • If DrG° <0 then ln(K) >0 so product is favored • If DrG° >0 then ln(K) <0 so reactant is favored

  6. Generalization of Above • If you have a number of products and reactants (all ideal gases), Dalton’s Law tells you PiV = vi RT where vi is the number of moles of the ith component (stoichiometric factor in the equation) • For each component, the Gibbs energy difference Gi -Gi° = RT ln(Pi) • At constant temperature, dG =VdP =nRTd lnP so • DRG -DRG° = RTS vi ln(Pi) and @ equilibrium DRG =0 • Remember DRGi° = DRGi° (products ) - DRGi° (reactants) = RT(S vi ln(Pi(prod) - S vi ln(Pi(reactants) ) • Thus -DRG° = RTS vi ln(Pi)equilibrium or -DRG° /RT = S vi ln(Pi) equilibrium • DRG° is a constant at t=constant so exp(-DRG° /RT ) = K • Further, S vi ln(Pi) = ln(P Pi vi )equilibrium • So K = P Pi vi • vi is positive for products and negative for reactants • In terms of activities (aA = pA/p*A), K =( P ai vi )equilibrium

  7. Notes on Generalization • For pure solids and liquds a=1 so they don’t contribute to the reaction quotient (or K) • For ideal gases (aA = pA/p*A), but for real gases (aA = fA/p*A), where fA is the fugacity • K is a function of temperature and DRG° is also a function of temperature • K is independent of total pressure and variation in partial pressures • Pressures varied by changing proportions of reactants and products consistent with K =( P ai vi )equilibrium • K is called the thermodynamic equilibrium constant • Like activities it is dimensionless • You can approximate K by replacing fugacities by partial pressures or molalities or molar concentrations • Poor approximation in electrolyte solutions or in conc. solutions

  8. Application to a Gas Phase Equation - Formation of NO @ 25°C • Reaction: N2 (g) + O2 (g) ˛ 2NO(g) • Data: DGf°(N2 ) = 0; DGf°(O2 ) = 0; DGf°(NO ) = +86.6 kJ/mol • DRG° = 2(86.6 kJ/mol) - 0 - 0 = 173.2 kJ/mol • K= exp(-DRG° /RT ) = exp (-173.2 kJ/mol/(2.48 kJ/mol)) • K = 4.67 x 10-31 • 4.67 x 10-31 = a2NO/ aN2 aO2 = (fNO p°)2/ (fN2) (fO2) • At low pressures fA =pA ,so K= 4.67 x 10-31 = (pNO p°)2/ (pN2) (pO2) • At atmospheric pressure pNO2 =(4.67 x 10-31 )(0.78084 x0.2094) = 2.86 x 10-31 atm; pNO = 1.69 x 10-15 atm

  9. Degree of Dissociation (Self Test 9.2) • For the equation, H2O(g), ˛ H2 (g) + 1/2O2 (g), what is the mole fraction of oxygen resulting from passing steam through a tube at 2000 Kif the Gibbs energy is 135.2 kJ/mol and P = 200kPa? K = exp(-DRG°/RT ) = exp ((-135.2 kJ/mol)/(8.3145*2000 J/mol)) =exp(-8.13)=2.945 x 10-4 If x moles of water dissociate then fraction left is 1-x, fraction of hydrogen = x and fraction of oxygen=x/2 Total pressure = {(1-x) + x +x/2}p = {1 + 1/2x} [p=200 kPa] Partial pressures :water = (1-x)p/(1+1/2x); H2 = xp/(1+1/2x); O2 = 1/2xp/(1+1/2x) K = {(xp/(1+1/2x)( 1/2xp/(1+1/2x))0.5)/[ (1-x)p/(1+1/2x)]; if 1>>x this simplifies to K = (1/2) 0.5(xp) 1.5 /p = (p/2) 0.5(x) 1.5 = 2.945 x 10-4 Substituting for p (200 kPa/105 Pa)and solving for x , (x) 1.5 = 2.945 x 10-4 ;so x = 0.004426 Mole fraction of oxygen = x/2= 0.00221

  10. Equilibrium Constant & Activity Coefficients • Recall that the activity, a is the product of the activity coefficient, g, and mole fraction, x or the activity coefficient, g, and concentration, b • Thus for A + B˛ C + D K = (gCaC)(gDaD )/(gAaA)(gBaB) or • K = (gCgD)/ (gAgB) ( aCaD )/ (aAaB)= Kg KB • For dilute solutions, Kg = 1 and K is approximately KB

  11. A Statistical View of Equilibrium • All molecules have energy levels corresponding to electronic vibrational and rotational energies • The total internal energy is the sum of the vibrational, rotational and electronic energies, eJ = ev + er + ee and the partition functions are also additive, zJ = zv + zr + ze • Partition function is the denominator in the Boltzman distribution • Boltzman distribution is the population distribution between states. • For a simple reaction A-> B, the equilibrium constant is the sum of the probabilities that the probabilities that the system be found at in one of the energy levels of B divided by the sum of the probabilities that it will be found in one of the states of A or K = z°B / z°B exp (-DeO/kT), where DeO is the difference between the lowest states of A and B

  12. Statistical View (cont.) • K = z°B / z°B exp (-DeO/kT), • If the spacing of energy levels the same, then the dominate species has the lower energy levels K dominated by DeO • Enthalpy dominates • If the spacing of the levels is higher for one species than the other (a greater higher density of states), the partition function will be greater and it will dominate K • Entropy donimates

  13. Example • Consider an endothermic reaction Na2˛ 2Na • DeO = 0.73 eV and exp (-DeO/kT) = 2.09 x 10-4 • Energetically equilibrium would be toward reactants • But K =2.4248 • Ground state of Na2 (A)is a singlet whereas ground state of atomic Na(B) is a doublet • Atomic Na has two nearly superimposed states and a much higher statistical weight

  14. Le Chatelier-Braun Principle • Henry Le Chatelier (1888) and F. Braun (1887) “Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation in the opposite direction” H. Le Chatelier or If a system in equilibrium is perturbed by subjecting it to a small variation in one of the variables that define the equilibrium, it will tend to return to an equilibrium state , which is usually somewhat different than the initial state

  15. Proof of Le Chatelier-Braun • Recall, dG = -SdT + Vdp + (S viµi) d x • Define the affinity, A , as (dG/dx ) p,T • At equilibrium, -A = 0 = S viµI (recall T and p are constant) • Now -d A = d (dG/dx ) p,T or from above • -d A = - (dS/dx ) p,T dT + (dV/dx p,T )dp + (d2G/d2x ) p,T d x • For all equilibrium states -d A = d (dG/dx ) p,T = 0 or - (dS/dx ) p,T dT + (dV/dx p,T )dp + - (d2G/d2x ) p,T d x = 0

  16. Proof of Le Chatelier-Braun(cont.) • At constant T, (dxe / dP)T = -[(dV/dx ) p,T ]/[(d2G/d2x ) p,T ], • xe is the equilibrium extent of reaction • (d2G/d2x ) p,T >0 at stable equilibrium (definition of stable equilibrium) so if P is increased at constant T, the extent of reaction increases in the direction that the volume of the system is decreased at constant T and P • At constant p, (dxe / dT) p = [(dS/dx ) p,T ]/[(d2G/d2x ) p,T ]= [T(dq/dx ) p,T]/[(d2G/d2x ) p,T ] dq is the reversible work and xe is the equilibrium extent of reaction • (d2G/d2x ) p,T >0 at stable equilibrium (definition of stable equilibrium) so if T is increased at constant p, the extent of reaction increases in the direction that heat is absorbed by the system at constant T and P

  17. Response of Equilibria to Pressure • Recall, K depends on DrG° • DrG° is defined at a standard presure, hence is independent of pressure so (dK/dp)T = 0 • Just because the constant is independent of pressure doesn’t mean the composition of gases is • Effect of additional inert gas - none because partial pressures unchanged (assumes ideality • Compression (changing volume): • Le Chatelier’s Principle says reaction will adjust to minimize pressure increase

  18. Response of Equilibria to Pressure Example: A -> 2B; K = pB2/pAp° • On compression [A] will increase to compensate for pressure change • If extent of reaction is a then we start with n moles the A = (1- a)n and b=2 a n pb • Mole fraction of A, xA = (1- a )n/((1- a)n + 2 a n) = (1- a )/(1+ a) • Mole fraction of B, xA = 2a /(1+ a) • K = [(2a /(1+ a))p]2/(1- a )/(1+ a)p = p[(4a2/(1+ a)(1- a )] or K = 4a2 p/(1- a 2 )or 4a2 p + K a 2 - K = (4p + K) a 2 - K = 0 • Note : p = p/p° • Solving for a 2 = K/ (4p + K) = 1/(1 + 4p/K) or a =1/(1 + 4p/K)1/2 • This says as p increases then the extent of reaction decreases

  19. Example: 3H2 + N2-> 2NH2 K = pNH32/p2H2pN2 K = xNH32p2/x3H2p3xN2 p where pA =xAp K = [xNH32 /x3H2 xN2 ] [p2/p 4 ] = [xNH32 /x3H2 xN2 ] [p2/p 4 ] K = KX 1/ p2 where, KX = [xNH32 /x3H2 xN2 ] or KXa p2 • Since K is constant, if p increases must increase as the square • Self test 9.3: If there is a ten-fold increase in pressure, what will be the effect on the constituents • KXa p2 so a ten fold increase in pressure will increase KX by 100 • This means that the products must increase or the reactants must decrease • Le Chatleier’s principle tells you that by inspection since reactants have the greatest contribution to the total pressure

  20. Effect of Temperature on Equilibrium -DG/RT = lnK so dlnK /dT =- (1/R)(d(DG /T)/dT) Recall Gibbs-Helmhotz equation (Ch. 5): d(DG /T)/dT = [d(DG /T)/d(1/T)][d(1/T)/dT] d(DG /T)/dT = [d(DG /T)/d(1/T)][1/T2] Now, DG = DH - TDS or DG/T = DH/T - DS d(DG /T)/dT = [d(DH /T -DS)/d(1/T)][1/T2] = DH /T2 (G-H eqn) Thus, d(lnK) /dT = DH /RT2van’t Hoff Equation • Because d(1/T)/dT = -1/T2 ,dT=-T2 d(1/T) d(lnK) /d(1/T) = -DH /Rvan’t Hoff Equation (alt. Form) • Says that plot of K vs. 1/T is a straight line with a slope = -DH /R • Can be used to estimate reaction enthalpy (see example 9.3)

  21. d(lnK) /d(1/T) = -DH /R • If reaction is exothermic DH <0 slope of ln(K) vs. 1/T is positive (>0) • As T increases 1/T decreases so as T increases thus K gets smaller • Increase in temperature favors reactants (le Chatelier) • If reaction is endothermic DH >0 slope of ln(K) vs. 1/T is negative (>0) • As T increases 1/T decreases so as T increases thus K gets larger • Increase in temperature favors products(le Chatelier)

  22. Statistical Mechanical Explanation • As T increases population of higher energy states is increased at the expense of lower states • For an endothermic reaction product states higher than reactants to it shifts to products • For an exothermic reaction, reactant states higher than products so reaction shifts to them

  23. Estimating K at Different Temperatures • dlnK /d(1/T) = -DH /R • Integrating between 1/T1 and 1/T2 • lnK2 - lnK1 = -DH /R(1/T2 - 1/T1 ) • This assumes H does not vary much over the temperature range • Self test 9.6 • For N2O4(g) ˛ 2NO2 (g) @ 298 K, K= 0.15 • What is value at 100°C (398K)? ln K1 = -1.897, 1/T1 =0.0033557, 1/T2 =0.00251256, (1/T2 - 1/T1 )= -0.00084314 DH =-9.16 + 2*33.18 = 57.2 kJ; DH/R = 6879.5 -DH /R(1/T2 - 1/T1 ) = -6879.5* (-0.00084314) = 5.8 Ln K2 = -1.897 + 5.8 = 3.903; K2 = 49.6

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